Let $R$ be a ring of subsets of a set $X$ and assume that $\mu : R → [0, ∞)$ is an additive function on
$R$, ie. $\mu(A ∪ B) = \mu(A) + \mu(B)$, whenever $A, B ∈ R$ with $A ∩ B = ∅$. Prove that
(i) $\mu$ is monotone; that is $\mu(A) ≤ \mu(B)$, whenever $A, B ∈ R$ and $A ⊆ B$;
(ii) $\mu$ is finitely subadditive.
Can someone give me any hints on how to go about starting this exercise, please?
Best Answer
$(i)$ Let $R\ni A\subset B\in R$ then $\mu (B\setminus A) \geq 0 $ and hence $$\mu (B) =\mu (A\cup (B\setminus A))=\mu (A) +\mu (B\setminus A) \geq \mu (A)$$
$(ii) $ Let $A_1 ,A_2 , .. , A_n \in R$ . Define $B_1 =A_1 ,$ $B_k = A_k\setminus \bigcup_{j=1}^{k-1} B_j. $ Then $$\bigcup_{j=1}^{n} B_j =\bigcup_{j=1}^{n} A_j$$ and $$B_j \subset A_j\mbox {and } B_j \cap B_i =\emptyset$$ hence $$\mu \left(\bigcup_{j=1}^{n} A_j \right) =\mu \left(\bigcup_{j=1}^{n} B_j \right) = \sum_{j=1}^{n} \mu (B_j )\leq \sum_{j=1}^{n} \mu (A_j )$$